Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
map<int,int>mat;
int n,p[10005],vis[10005];
ll a[505][505];
ll gcd(ll x,ll y){
	if (!y)return x;
	return gcd(y,x%y);
}
ll lcm(ll x,ll y){
	if ((!x)||(!y))return x+y;
	return x/gcd(x,y)*y;
}
int nex(int x){
	if (!mat[x])mat[x]=p[p[0]--];
	return mat[x];
}
int main(){
	scanf("%d",&n);
	if (n==2){
		printf("4 7\n23 10\n");
		return 0;
	}
	for(int i=2;i<=10000;i++){
		if (!vis[i])p[++p[0]]=i;
		for(int j=1;(j<=p[0])&&(i*p[j]<=10000);j++){
			vis[i*p[j]]=1;
			if (i%p[j]==0)break;
		}
	}
	for(int i=1;i<=p[0]/2;i++)swap(p[i],p[p[0]-i+1]);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if ((i+j)%2==0)a[i][j]=nex(i+j);
	mat.clear();
	for(int i=n;i;i--)
		for(int j=n;j;j--)
			if ((i+j)%2==0)a[i][j]*=nex(i-j);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if ((i+j)%2)a[i][j]=lcm(lcm(a[i-1][j],a[i][j-1]),lcm(a[i+1][j],a[i][j+1]))+1;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)printf("%lld ",a[i][j]);
		printf("\n");
	}
}